Construction Strikes Again!

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Suppose you know the construction to find the midpoint of a line segment, as described in the previous post.  By now you realize that as a by-product, we also construct a line CD perpendicular to the original segment AB.  So, without making too much of a fuss, we know how to construct a perpendicular bisector of a given segment.

I was going to go on to the next construction, which was simply to construct a perpendicular to a given line, but wait; this is America, and we're not allowed to teach anything boring, just because it is good information.  (Oh sinful generation...)  Ok, so here's something a little jazzier, for instant gratification.  (Who knew we had that in geometry, eh?)

It turneth out (as King James might have said), that if you've got a line segment --call it, say, oh, I don't know ... how about AB?  Okay-- well, if you make the perpendicular bisector, every point on it (on the perpendicular bisector, that is, or bisectoris rectus, as Augustus Caesar would most certainly not have said, but I could be wrong) is equi-distant from A and B!

This brings us to an interesting result.  If you pick any triangle at random, it so happens that it is possible to draw a circle through the three points of the triangle!  Not everybody is as excited about this as we math folks are, but this is a fundamental fact of Euclidean Geometry (the geometry that seems to describe the world in which we live pretty accurately, and where the circumference of a circle of diameter 1 is Pi, and so on).

How one constructs this circle --called the circumcircle-- is interesting, and quite an easy application of perpendicular bisectors.  Why it happens to exist is another matter entirely, and at this moment I'm not totally feeling brave enough to embark on an explanation.  Maybe at a later time.

So, as indicated before, here at right (dextra) is a step-by-step diagram about how it's done: you just construct perpendicular bisectors of any two sides of your triangle; miraculously, if the third perpendicular bisector were to be drawn, all three of them would cross at the same point, and this is the center of the fabulous circle, which is all you really need.  Just for excitement, I show all the trillions of circles we draw to get the perpendicular bisectors; otherwise the sketch would look pretty dull.

Proof

Okay; it's tomorrow, and I've got a proof ready for you.  For this proof, we construct the first two perpendicular bisectors, mark where they intersect as O, and then, join O to the three vertices, A, B and C.

Since the two blue segments are perpendicular bisectors, the two yellow-filled triangles are congruent, so the big yellow triangle they form together is isosceles, which is what makes it possible to draw the circle in the first place, actually.

Similarly, the two green triangles are congruent, and the big green triangle is isosceles.

Now we have to show that if we were to make the third perpendicular bisector, it would go through O too.

How does one do such a thing?  We just mark the midpoint, M of BC, and join it to O.  Now if only OM turns out to be a perpendicular to BC, our point would have been proved.

But it is possible to show that the two purple triangles, OMB and OMC are congruent too (SSS, or "side-side-side", since we know already that OB and OC are congruent, from the earlier reasoning.)  So OM is in fact a perpendicular.  There can be only one perpendicular to a line at a given point, so OM is the perpendicular bisector of BC.  So that's how the whole thing works, QED.